Major League Baseball has always been particularly inept when it comes to their tiebreaking rules. Amazingly, for the first few years of the wild card, no rules were in place at all as to what happens when various multi-way ties occur in division and wild card races. To the outside, it seemed that MLB was making up the rules as it went along. Finally in 2003 some rules were codified as to what happens in the event of multi-way ties. Being the end of the baseball season, I found it an appropriate time to discuss baseball's uneven tiebreaking rules. Believe it or not, the current rules for most multi-way ties are quite inequitable and do not give all teams even close to a fair chance at the playoffs. Thankfully no multi-way ties have ever occurred, but were it to happen, baseball's current rules don't allow for the fairest of all possible outcomes. Here I'll walk through various tie scenarios, break down MLB's current rules, and propose better, fairer solutions to many multi-way ties.

Three-Way Tie for Wild Card (or Division)

One of baseball's most common multi-tie scenarios (and I use "common" loosely since it has never occurred) is the three-way tie for a wild card or division title. In this case, three teams are fighting for one playoff spot. You're probably already aware how this works - it's what baseball calls its A/B/C tiebreaker. Team A hosts Team B and the winner of that game hosts Team C. The winner of that game is the champion. In a hypothetical example, the Mets would host the Cubs and the winner of that game would host the Giants.

The Flaw: The solution indeed resolves the issue in just two games, but does so very inequitably. The Mets and Cubs are forced to win two games, while the Giants must win just one. Calculating the probabilities, I've assumed that the home team has a 56% chance of winning each game (bumped up slightly from the usual 54% because of playoff crowds and the increased travel hassle for unexpected tiebreaking games). When we do so for this scenario, the Giants (playing just one game on the road) have a 44% chance to win, while the Cubs (playing one on the road and then one at home) have just a 25% chance to win. A good tiebreaker gives each team an equal chance for victory, and the current tiebreaking rules do a terrible job, giving one team nearly double the advantage over the another. If I am the Cubs, I am going to be very angry with this set-up.

The Solution: While there's no way to remedy inequity this in just two games, I believe it would be worth just one more game to more fairly determine the outcome. In this solution, the champion would have to win two games to advance. In the example, the Giants would host the Cubs and the winner would then go to New York for a potential double-header (or games on back-to-back days if you must). To advance, the Mets would be forced to win both games, while the Cubs or Giants would need just one win in New York to advance. In this scenario, the probabilities are more fairly distributed at 38%-31%-30% instead of the current 44%-31%-25% breakdown. The solution would require only one potential extra game, no extra days, and would require no extra travel by any team. The probability chart below shows the improvement and the flow chart shows how the tiebreaker would work in a graphical form.

Three-Way Tie for the Division and the Wild Card

Another common three-way tie scenario is one in which three teams from one division are tied for both the division and the wild card. In this case, three equal teams are fighting for two playoff spots. I actually haven't come across this codified anywhere, but I presume MLB would again employ the A/B/C tiebreaker albeit in a slightly different form. In the case of a Dodgers-Giants-Rockies tie, LA would host SF and the loser would play the Rockies in Colorado. The winners of each game would go to the playoffs.

The Flaw: In this case, the Rockies are clearly getting the short end of the stick. They are in a must-win situation while LA and SF must win just one of two. Probability-wise, the Rockies have just a 56% chance to advance, with the Dodgers' chance is 75%. This is seriously unfair and is no way to decide the season.

The Solution: The way to fix this is much the same as the scenario above. By extending the tiebreak by just one potential extra game at the same site, we can even the odds. In the improved scenario, the Dodgers host the Giants. The winner advances while the loser goes home to host the Rockies for two games. If the Rockies win just one game, they advance, while the loser of the LA-SF matchup must sweep both games in order to salvage the wild card. This plan, seen below, provides a far more equitable 70%-69%-61% distribution of probabilities rather than the currently unfair 75%-68%-56%. Again, this solution requires just one potential extra game with no extra travel to decide things more fairly.

Two Tied for Division and One Other Tied for the Wild Card

This scenario, in which two teams are tied for the division and also are tied with one other team for a wild card spot, was the doomsday scenario for MLB for the first few years of the wild card era. In a theoretical LA-SF-CHC scenario, MLB would have had LA and SF play for the division title, while the Cubs were gifted the wild card. The theoretical ideal playoff odds should have LA and SF with a 75% chance to make the playoffs, and the Cubs at a 50% chance. However, this resulted in a 56%-44%-100% split instead - a major error. Thankfully, Selig and Co. have corrected this so that the LA-SF loser would play against the Cubs for the wild card.

The Flaw: This new solution is by far a better one, but is not optimal by any means. With the home field advantage determined by head-to-head records, it's possible for one team to gain a far larger advantage than it should. The current tiebreaker could have a LA having the home field in both games, while SF could be on the road in both games. This would lead to LA having an 81% chance to advance, while SF's probability would be just 69%. Other unfair possibilities include having the Cubs with home field against either team, giving them a 56% chance of advancing rather than 50%.

The Solution: This can be fixed easily enough, and unlike the above prescriptions, can be done without any additional games. The solution is simple. If one team has the home field in the first game, they should not have it in the second game, and vice-versa. This leads to a more equitable 75%-75%-49% split, nearly a perfect match to the theoretical ideal probabilities. This scenario can happen under the current rules if the head-to-head matchups fall correctly, but MLB should codify it so that it is always the case. The charts below show the difference in probabilities (with the scenario discussed above as the "current" scenario, though there are other ways the probabilities could fall under current rule as well) as well as the way the improved tiebreak would work.

Four Tied for Wild Card (or Division)

Getting into unlikelier four team scenarios, we also see flaws in the MLB's method. This one is where four teams are tied and fighting for just one playoff spot. According to the MLB rules, "Club"A" shall play Club"B" at the ballpark of Club "A" and Club "C" shall play Club "D" at the ballpark of Club"C". The following day, the winner of these games shall play one game, at the ballpark of Club "A" or Club "B," whichever has won the game between the two."

The Flaw: The MLB solution sounds fine enough, but they again blow it with the home field advantage. Under the MLB rule, in a LA-COL-SF-SD tie, where COL hosts SD and SF hosts LA, if the Rockies and Dodgers win their games, Colorado would host it's second game while the Dodgers would go on the road for the second time in a row. There's no reason why Colorado should have the advantage of getting both home games, while the Dodgers suffer twice on the road. The flaw results in a 31% chance of advancing for Colorado, while the Dodgers' probability is just 19%.

The Solution: Again the solution is as simple as allowing the road team in the first game to play at home for the second game and vice-versa. Of course, if both home teams (or both road teams) win their respective games, then one of them must be at home again, giving them an advantage, but this can hardly be avoided. The improved tiebreaker allows for a 28%-25%-25%-22% split rather than the current 31%-25%-25%-19% probabilities. Since this requires no extra games or travel, this fix is a no-brainer. The probability charts and flow charts are below.

Four Tied for Division and the Wild Card

When four teams are tied for the division and also tied for the wild card, this is the one scenario where MLB gets it right. Since four teams are fighting for two spots, the simplest and best solution is to play two games, with the winners of each advancing. In the NL West scenario, where LA, SF, COL, and SD are tied, the easiest solution is to have LA host COL and SF host SD, where the winners advance.

The Flaw: The system is not perfect, since obviously the home teams in each of the games have the advantage, but this can't be improved without playing many more games (or playing at neutral sites, which would be awful).

The Solution: MLB's solution is a good one that is as fair as can be. Below are the probabilities and the (extremely simple) accompanying chart.

Three Tied for Division and One Additional Team Tied for the Wild Card

Another scenario where MLB's policies are egregious is where three teams are tied for the division lead and one additional team is tied for the wild card. Under current rule, the three teams tied for the division title play an A/B/C tiebreaker (A hosts B and the winner hosts C). Then the two losers play another A/B/C tiebreaker with the wild card team as the third team. In a hypothetical situation, the Dodgers, Giants, and Rockies would play an A/B/C tiebreaker. Then the two losers would play an A/B/C tiebreaker with the Cubs, who had an identical record as the other three teams.

The Flaw: There's nothing terribly wrong with the format, but when MLB does not control for who gets the advantage of being Team C, we can get some skewed probabilities. In the perfect world, the three NL West teams should have a 55.55% chance of advancing, while the Cubs should have a 33.33% chance of advancing. However, depending on who gets lucky or unlucky in being Team A, B, or C, these can be thrown off wildly. For instance, if the Giants get to be Team C (playing just one game on the road) for both A/B/C tiebreakers, then their probability to advance to the playoffs is 69% - far higher than it should be, while if the Rockies are Team B for both A/B/C tiebreakers then their chance to advance is just 43%. This is big disparity for two teams that should have the same chance of advancing.

Furthermore, in another scenario (not reflected in the table), the Rockies could be Team B for both tiebreakers, while the Cubs could be Team C for the second tiebreaker. This would actually give the Cubs a higher probability (44%) of advancing than the Rockies (43%), when in fact the Colorado should have a major advantage over Chicago.

Another oddity is that theoretically SF could advance by going 1-1 (losing as Team C in the first tiebreaker and winning as Team C in the second tiebreaker) while Colorado could be eliminated while going 2-2 (winning and then losing in both the first and second tiebreakers). This would be a very peculiar way to break a tie, since both teams' records would be essentially tied again after the extra-curricular play. I have a feeling the Rockies and their fans wouldn't be too pleased with this outcome.

The Solution: The solution of course is to control for who gets the A/B/C advantage in each round, so that one team cannot benefit by being Team C in both rounds. The improved scenario would work like this: This Giants would host the Rockies in the first tiebreaker. The winner would then host the Dodgers for the NL West title. In the second round, the Dodgers (if they had not won) would get the worst of it as Team B, while the loser of the SF-COL game would become Team C. This ensures that the Dodgers, who got the best draw in round 1, will get the worst draw in round 2. If the Dodgers end up winning the first tiebreaker, the winner of the COL-SF game gets to be Team C, while the loser becomes Team B. Meanwhile the Cubs are constantly fixed at Team A in the second round tiebreaker.

Confused? The chart below probably shows it more clearly than it can be explained in words. Overall the Dodgers, as Team C in the first tiebreaker and Team B in the second tiebreaker have a 58% chance of advancing - close to the ideal 55.6%. The Giants, who were home as Team A in the first tiebreaker, also have a 58% chance of advancing. The Rockies draw the short straw by being the road team in the first game and have a 53% chance of advancing. Meanwhile, the Cubs, fixed as Team A in the second tiebreaker, have a 31% chance of advancing - very close to the ideal 33.3%. Overall, the solution comes very close to matching the ideal probabilities - something that the current MLB rules do not do. This is a must fix.

Two Tied for the Division and Two Additional Teams Tied for the Wild Card

Another possible four team scenario is where two teams are tied for the division and two other teams are tied with them for a potential wild card spot. For example, say LA and Colorado are tied for first in the NL West, while the Cubs and Mets have identical records and are hoping for the wild card. Under current rules, the Rockies and Dodgers will play first. The winner advances while the loser plays an A/B/C tiebreaker with the Cubs and Mets.

The Flaw: This scenario is again fine, except for the fact that the same NL West team can get lucky (or unlucky) in both tiebreakers and gain an unfair advantage regarding their home field and whether they become Team A, B, or C. The theoretical odds should give LA and Colorado a 66.7% chance to advance, while the Cubs and Mets should have a 33.3% chance to advance. However, if LA plays at home in the first tiebreaker and also becomes Team C for the second tiebreaker, their odds increase to 75%. Meanwhile, Colorado, if they are unlucky enough to become Team B in the second tiebreaker, has just a 58% chance to advance. This disparity is unacceptable.

The Solution: Clearly the solution is to make sure that one team does not get the advantage in both tiebreakers. An improved method is to have Colorado host LA just as before. If Colorado wins, LA moves on to the wild card tiebreaker, but since they were on the road before, they get to be Team C in the second tiebreaker. Likewise, if LA wins, the Rockies must be Team B in the wild card tiebreaker, since they had home field the first time. The Mets are fixed as Team A, while the Cubs become either Team B or Team C depending on the outcome of the LA-COL game. The new solution creates a probability breakdown of 69%-67%-33%-31%, which is much better than the potential 75%-57%-38%-28% scenario which is possible under current rule. Below is a chart of potential probabilities and a flow chart of how the improved tiebreaker would work.

Two Teams Tied for Two Different Divisions and the Wild Card

The final four team scenario we'll look at is if two teams are tied for one division, another two teams are tied for another division, and all four teams are tied for the wild card. In a hypothetical example, LA is tied for Colorado in the West and the Cubs and Cardinals are tied for the Central, with all four teams also eligible for the wild card. Under current rule, both divisions will be decided by one-game playoffs with the losers playing for the wild card.

The Flaw: As with the other scenarios, one team can gain a double advantage by being the host of both games, while another team may be forced to go on the road for each game. If Colorado hosted LA, and the Cardinals hosted the Cubs, it's possible that LA would end up going to St. Louis, giving the Cards two home games and LA two road games. All teams should theoretically have the same probability of advancing (75%), so this clearly is unfair to the Dodgers.

The Solution: The scenario above can be rectified by forcing the Cardinals to go to LA rather than vice-versa. Like the other tiebreaking solutions, teams with the advantage in the first tiebreaker should not be given the advantage again where possible. The charts below show the improved solution, where we see more equitable probabilities.

Conclusion

As you can see, there is much room for improvement in MLB's tiebreaking procedures (I haven't explained the probability calculations in detail, but feel free to ask questions in the comments). While at least now they have some procedures in place, they are not the fairest solutions to the ties. Tied races are one of baseball's rarest and most exciting events, and when a three-way or four-way tie inevitably occurs, baseball should make sure it has the procedures down right (I've also developed some 5 and 6 way tiebreaking procedures, but that will be saved for a later post if there's interest). Implementing the procedures just outlined can go a long way towards making things more fair and avoiding controversy in the future.

Comments

Additionally, you could cut into the home field advantage by giving the team not playing in their home field the last at bat. Might be odd for the fans, but batting last is part of the home field advantage. Flip a coin, let the winner choose home field or home batting (They'd always choose home field for the $).

Posted by: Brian at September 14, 2009 9:02 AM

The best solution that would solve most of the problems above: the abolishment of the wildcard.

Posted by: daniel at September 14, 2009 9:34 AM

Sky - That is some pretty indepth stuff and actually good. What is your backround?

Does head to head not currently count at all?

in 2005 The Yankees and Soxs tied and Dodgers and Padres in 2006 tied. How do they figure who is the Division winner and who is the wild card? In 05 the Yankees won head to head and the division in 06 the Padres won head to head but the Dodgers won the division.

Another curious aspect of this is the nature of the 163rd game itself. For the sake of record keeping, I believe, it is considered a regular season game. If you define the wild card winner as the team that has the highest winning percentage in the regular season (outside of the division champs, of course), then the loser of the 163rd play-in game for a division tie-breaker would no longer be eligible for the wild card tie-breaker (as their regular season winning percentage is lower).

Posted by: Pat at September 14, 2009 11:40 AM

Pat, that used to be MLB's philosophy before they thankfully changed the unfair rules.

Posted by: Sky Andrecheck at September 14, 2009 11:51 AM

Sky, theres a better solution to the 3-way tie for wildcard. Homefield is determined at random. All 3 teams play 2 games on that field.
NYM @ SF
CHC @ SF
The winners of these two teams play the next day in San Francisco. That eliminates unnecessary travel, and eliminates the homefield advantage of the rubber game.

Posted by: cdm at September 14, 2009 12:19 PM

cdm,

Yes, a lot of these can be improved further if you allow games at neutral fields. Personally though, I think a tiebreaker at a neutral field would stink, and I would bet MLB owners would agree, so I didn't allow them in my analysis.